24 сент. 2013 г. · It is continuous. However, it is not uniformly continuous. Suppose it were; then for every ϵ, there exists a δ for which |x−y|<δ⟹|x2−y2|<ϵ.

25 июл. 2015 г. · x2 is uniformly continuous on [0,1] because continuous functions are always uniformly continuous on a compact set.

16 нояб. 2020 г. · Your intuition as to why it's not uniformly continuous is correct, but your proof doesn't make any sense. Start with the definition of not ...

25 июн. 2023 г. · To prove that x2 is not uniformly continuous, you need to show that there exists an ε>0 such that for every δ>0, there exists x,y such that |x−y ...

17 авг. 2020 г. · It's certainly uniformly continuous on [0,1]. In general, a continuous function will always be uniformly continuous on a compact set.

19 авг. 2020 г. · Proving that x2 is not uniformly continuous ... we find |x−y|<δ, yet |f(x)−f(y)|≥ϵ. Hence the ϵ−δ definition of uniform continuity is negated and ...

30 окт. 2019 г. · If you can find a modulus that does not depend on x0, as in if you vary x0, then the value of N doesn't change, then your function is uniformly ...

27 февр. 2015 г. · Yes, x2 is uniformly continuous on any closed interval [a,b] but not uniformly continuous on the real line. – WillO. Commented Feb 27, 2015 at ...

26 июн. 2015 г. · Show that f(x)=x2 is not uniformly continuous on [0,∞) If we replace y=2x, then ∣x−2x∣=∣−x∣=x<δ. So, that holds. Now, ∣f(x)−f(y)∣=∣x ...

29 янв. 2019 г. · x^2+x is between x^2 and (x+1)^2 . It is well-known x^2 is uniformlycontinuous and thus by graph-similarity (x+1)^2 is uniformly continuous.